The proof goes via the Atiyah-Hirzebruch spectral sequence, the claim being that the AHSS degenerates at the $E_2$-page $E_2^{p,q} \cong E^{\ast}[x] / (x^{n+1})$. I don't understand why the differentials have to vanish. Could somebody explain this to me in detail? Shouldn't be difficult, but I'm not familiar with the AHSS and don't see it.

2 Answers
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If $E$ is an even cohomology theory (i.e., $E^i(*) = 0$ for odd $i$), then the objects in the $E_2$ page are only nonzero in degree $(p,q)$ with $p$ and $q$ even. In particular, the total degree is even. The differentials of a spectral sequence increment total degree by one, so they only hit entries of degree $(p,q)$ where $p+q$ is odd. Those entries are zero, so the differentials vanish.

Only a small bit, though. The parity problem is concentrated in one variable under Adams grading, but it's still fundamentally a parity proof. Charles's argument is strictly more general, and not much longer.
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S. Carnahan♦May 20 '10 at 5:56

I'll augment Scott's answer, and point out that you don't even need $E$ to be even. The class $x\in E^* CP^n$ is detected by an element $\bar{x}$ in the $E_2$-term. Because you know the class $x$ exists, $\bar{x}$ survives to $E_\infty$; that is, $d_r(\bar{x})=0$ for all $r\geq2$.

The differentials are derivations of $E_*$-algebras, so every element of the subring of $E_2$ generated by $E^*$ and $\bar{x}$ survives to $E_\infty$. But this is the whole $E_2$-term, so there are no non-trivial differentials.